Solving the Cubic Equation

Solving the Cubic Equation Problems from the History of Mathematics Lecture 9 | February 21, 2018 Brown University A Brief History of Cubic Equations The study of cubic equations can be traced back to the ancient Egyptians and Greeks. This is easily seen in Greece and appears in their attempts to duplicate the cube and trisect arbitrary angles. As reported by Eratosthenes, Hippocrates of Chios (c. 430 BC) reduced duplicating the cube to a problem in mean proportionals: a : b = b : c = c : 2a: This idea would later inspire Menaechmus' work (c. 350 BC) in conics. The Greeks were aware of methods to solve certain cubic equations using intersecting conics,1 but did not consider general cubic equations because their framework was too influenced by geometry. 1One solution is due to Apollonius (c. 250-175 BC) and uses a hypoerbola. 1 A Brief History of Cubic Equations The Greek method of intersecting conics was completed by Islamic mathematicians. For example, Omar Khayyam (1048-1131) gave 1. examples of cubic equations with more than one solution 2. a conjecture that cubics could not be solved with ruler and compass 3. geometric solutions to all forms of cubics using intersecting conics While the techniques were geometric, the motivation was numeric: the goal in mind was to numerically approximate the roots of cubics.2 These ideas were taken back to Europe by Fibonacci. In one example, Fibonacci computed the positive solution to x3 + 2x2 + 10x = 20 to 8 decimal places (although he gives his solution in sexagesimal). 2Work done resembles Ruffini’s method. 2 Cubics in Renaissance Italy Cubics in Renaissance Italy The first real step towards the full algebraic solution of the cubic came from an Italian mathematician named Scipione del Ferro (1465-1526), who solved the depressed cubic3 x3 + px = q: Scipione del Ferro kept his method secret until right before his death. This secrecy is likely a result of the mathematical culture of the era, in which losers in a mathematical contest might lose funding or position. Scipione del Ferro passed his solution onto Antonio Fiore. 3It would have been assumed that p; q ≥ 0. 3 Cubics in Renaissance Italy In 1530, another Italian named Tartaglia claimed a solution to two cubics presented by Zuanne da Coi. This led to a contest between Tartaglia and Fiore, in which Tartaglia submitted cubics of the form x3 + px2 = q while Fiore submitted those of the form x3 + px = q. As Fiore could only solve depressed cubics, Tartaglia won.4 4It's not clear how Tartaglia came to his more general solution. 4 Cubics in Renaissance Italy Tartaglia thus caught the attention of Gerolamo Cardano (1501-1576), who convinced Tartaglia to divulge his method in 1539 under an NDA. When Cardano learned that Tartaglia's work was known to del Ferro, he broke the agreement and published Ars Magna in 1545. Tartaglia then challenged Cardano, who declined. The challenge eventually passed to Cardano's student Ludovico Ferrari (1522-1565), who understood cubics better and bested Tartaglia at his own game. 5 Solution to the Depressed Cubic One Solution to the Depressed Cubic The key to deriving any of the cubic formulas is to find a substitution which reduces the problem to a quadratic equation. Here is one: 1. (α − β)3 = α3 − 3α2β + 3αβ2 − β3 2. (α − β)3 − 3αβ(α − β) = α3 − β3 3. α − β is a root of the cubic x3 − 3αβx = α3 − β3 4. Given a cubic x3 − px = q, we just need to find α; β such that 3αβ = p and α3 − β3 = q: 5. Substitution gives p 3 α3 − = q; 3α which (after multiplying by α3) is a quadratic polynomial in α3. 6 One Solution to the Depressed Cubic Solving the quadratic polynomial from before gives r q p3 q 2 α3 = ± + : 2 3 2 Note that we can move from the case x3 = px + q to x3 + px = q using the substitution p 7! −p. If we do, the term in the square root remains positive only when (q=2)2 > −(−p=3)3. This case was referred to as the casus irreducibilis in Ars Magna. This is perhaps because Cardano thought it coincided with the lack of a positive real root.5 5It doesn't. 7 The Introduction of Complex Numbers Complex Numbers Cardano was aware that his algebra supported the casus irreducibilis but clearly disliked the idea of working with negative square-roots. Example (Ars Magna, ch. 37): Find two numbers which sum to 10 p p and multiply to 40. (Cardano produces 5 + −15 and 5 − −15 and describes the consideration as `mental torture.') 8 Complex Numbers The casus irreducibilis would not be furthered until 1572, when Rafael Bombelli published l'Algebra and discussed the cubic in full. He considers the cubic x3 = 15x + 4, for which Cardano's formula gives q p q p x = 3 2 + −121 + 3 2 − −121: Bombelli observes that x = 4 is a root and explains how the formula does in fact produce this. His justification involves setting q p 3 2 + −121 = a + bi and solving for a = 2 and b = 1.6 While complex numbers were still far from accepted, Bombelli's work showed their potential promise. 6Here, it is convenient that Bombelli has introduced a notation for i. 9 Questions? 9.

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